3 Answers
3

Since $a_n=\binom{a+n}{n}^{-1}$, try to represent it using the difference of two sequence elements when replacing $a$ by $a-1$ or $a+1$,
\begin{align}
\binom{a-1+n}{n}^{-1}-\binom{a+n}{n+1}^{-1}
&=
\frac{n!}{(a+n-1)...(a)(a-1)}-\frac{(n+1)!}{(a+n)...(a)(a-1)}\\[0.8em]
&=\frac{n!(a+n-(n+1))}{(a+n)...(a)(a-1)}=a_n
\end{align}
which allows you to examine this as telescoping series.

The difference of reciprocals of binomial polynomials of degree $a-1$ is proportional to the reciprocal of a binomial polynomial of degree $a$; to be precise,
$$
\frac1{\binom{n+a}{a}}=\frac{a}{a-1}\left[\frac1{\binom{n+a-1}{a-1}}-\frac1{\binom{n+a}{a-1}}\right]\tag{1}
$$
If we sum in $n$, the right side of $(1)$ telescopes. Thus, for $a\gt1$,
$$
\begin{align}
\sum_{n=1}^\infty\frac1{\binom{n+a}{a}}
&=\frac{a}{a-1}\frac1{\binom{a}{a-1}}\\
&=\frac1{a-1}\tag{2}
\end{align}
$$
For $a=1$, the series diverges since
$$
\sum_{n=1}^\infty\frac1{\binom{n+1}{1}}=\sum_{n=1}^\infty\frac1{n+1}\tag{3}
$$
diverges.